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2 years ago

WILL MARK BRAINLIEST!!!

Mathematics

2 answers:

Sergeu [11.5K]2 years ago

3 0

Answer:

The answer is the bottom graph on the left. :)

kondor19780726 [428]2 years ago

3 0

Answer:

Bottom left corner.

Step-by-step explanation:

in order to solve this, you just have to evaluate the function when x=0 that means the intersect with Y-axis, so when h(x) = x^3 + 2x^2 - 11x - 12 is evaluated for x=0

y=-12

So the point would be (0,-12)

From the graphs there are two graphs that match this description, now we just have to evaluate the intersect with X, in the bottom left it says that when y=0 x=-4

So we evaluate h(x) = x^3 + 2x^2 - 11x - 12 to x=-4 to see if it equals 0:

$h(x) = x^3 + 2x^2 - 11x - 12\\h(x) = -4^3 + 2(-4)^2 - 11(-4) - 12\\h(x) = -64 +32 +44 - 12\\\\h(x)=0$

As you can see the function equals 0 that means that is the graph of h(x) = x^3 + 2x^2 - 11x - 12

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The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.512.512, point, 5 years; the stan

zaharov [31]

This question was not written properly

Complete Question

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.

Answer:

Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585

Step-by-step explanation:

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Mean is given in the question as: 12.5

Standard deviation : 2.4 years

We start by applying the first rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

μ - σ

12.5 -2.4

= 10.1

We apply the second rule

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

μ – 2σ

12.5 - 2 × 2.4

12.5 - 4.8

= 7.7

We apply the third rule

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

μ - 3σ

= 12.5 - 3(2.4)

= 12.5 - 7.2

= 5.3

From the above calculation , we can see that

5.3 years corresponds to one side of 99.7%

Hence,

100 - 99.7%/2 = 0.3%/2

= 0.15%

And 10.1 years corresponds to one side of 68%

Hence

100 - 68%/2 = 32%/2 = 16%

So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%

= 15.85%

Therefore, the probability of a lion living between 5.3 to 10. 1 years

is converted to decimal =

= 15.85/ 100

= 0.1585

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Which of the following could be the equation for a parabola that opens to the left with vertex (-17,2)?

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A sideways opening parabola is in the form $x= y^{2}$ , so we know from the process of elimination that it will either be b or c. Next we have to realize that if the parabola opens to the left it is a negative parabola, just like if a parabola opens upside down it is a negative parabola. So the one that has the negative out front is b.

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2 years ago

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Answer=1/24

1/3+5/8=
to solve for this, we need the denominators of the fractions to match.

The LCM of 3 and 8 is 24

1/3=8/24
5/8=15/24

8/24+15/24=23/24

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Of the 100 students enrolled at the dance studio, 50 take ballet and 24 take tap dancing. Eight of the students who take ballet

Soloha48 [4]

Answer:

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Step-by-step explanation:

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kozerog [31]

Answer:

(a) 0.06154

(b) 0.2389

(d) 2478

Step-by-step explanation:

probability density function of the time to failure of an electronic component in a copier (in hours) is

P(x) = 1/1076e^−x/1076

λ = 1/1076

A) A component lasts more than 3000 hours before failure:

P(x>3000) = 1 − e^−3000/1076

= 0.06154

B) A component fails in the interval from 1000 to 2000 hours:

P(1000>x>2000) =1 − e^−2000/1076 − 1 +e^−1000/1076 = e^−1000/1076 − e^−2000/1076 = 0.3948 − 0.1559

= 0.2389

C) A component fails before 1000 hours:

P(x<1000) = 0.6052

D) The number of hours at which 10% of all components have failed:

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